What is the equivalent capacitance of this formation?

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SUMMARY

The discussion focuses on calculating the equivalent capacitance of a circuit formation involving capacitors C1 (4.1 µF), C2 (6.86 µF), C3 (5.03 µF), and C4 (5.92 µF). Participants clarify that C2 and C3 are in series, leading to a combined capacitance C'. C1 and C' are then in parallel, followed by C" (the combination of C1 and C') being in series with C4. The importance of accurately identifying series and parallel connections based on component endpoints is emphasized for correct calculations.

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123lt
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http://<a href="http://imgur.com/UwulXqg"><img src="http://i.imgur.com/UwulXqg.png" title="source: imgur.com" /></a>[/PLAIN] 1. Homework Statement
Find equivalent capacitance of this formation from a to b, C1=4.1 uF, C2=6.86 uF, C3=5.03 uF, C4=5.92 uF

http://imgur.com/UwulXqg

Homework Equations


C series= (1/c)+...(1/c(x))=1/Ceq
C parallel= C+...C(x)=C eq

The Attempt at a Solution


I have a bit of trouble in these sort of more complicated formations. What I know for sure is that C2 and C3 are in series, therefore C'=(1/C2)+(1/C3). I'm guessing that C1 and C' are then parallel, and that (C1+C') and C4 are then in series. If someone could help explain how to figure out how to easily tell if something is parallel vs in series I would appreciate!
 
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123lt said:
If someone could help explain how to figure out how to easily tell if something is parallel vs in series I would appreciate!
Look at the ends of the components. If they are BOTH joined, then the components are in parallel.
IE. if component A has ends going to x and y, and component B has ends going to x and y, then A and B are parallel and their equivalent is connected to x and y.
If component A joins x to y, and component B joins y to z (and z is not directly connected to x) then A and B are in series. Their equivalent is connected from x to z. (BUT this only works if y is not connected to anything else!)

But be careful on your diagram : there are lots of connections!
 
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Merlin3189 said:
Look at the ends of the components. If they are BOTH joined, then the components are in parallel.
IE. if component A has ends going to x and y, and component B has ends going to x and y, then A and B are parallel and their equivalent is connected to x and y.
If component A joins x to y, and component B joins y to z (and z is not directly connected to x) then A and B are in series. Their equivalent is connected from x to z. (BUT this only works if y is not connected to anything else!)

But be careful on your diagram : there are lots of connections!
Therefore would C1 and C' be in series (making C") and from there C" and C4 are in series?
 
No. You are not taking account of the other wires/connections. Look at all the things a is joined to and at all the things b is joined to.
 
Merlin3189 said:
No. You are not taking account of the other wires/connections. Look at all the things a is joined to and at all the things b is joined to.
just to be clear, wires do not count do they?
Does that mean C and C' are parallel? Because then it would be like a is connected to the bottom side of C', making them parallel. And those together, C" would be parallel to C4 then? I'm sorry this is taking me a while to figure out.
 
No problem. Take your time.
Yes you have the right idea.
I sometimes find it helpful to redraw the circuit to make it clear. Here they gave all the capacitors sloping. People often find it easier to understand when they are horizontal or vertical. (In fact I redrew it to post here and the program I used would not let me put them sloping!)

When you say, "wires don't count", they don't count as components, but they are important - to show which bits are joined. You can move wires and lengthen or shorten them, so long as they still join the same components.
 
Great, I got the answer correct thank you for your help.
 
Just to show you what it looked like when I redrew it in Ltspice, I've uploaded it (temporarily)
It looks much easier when you see it like this, I think.

Edit- now you've seen it, I've deleted it to save space
 
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that makes a lot more sense, I will have to practice redrawing circuits to make them more clear! thank you again for your help.
 

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